32 CHAPTER l • COMPLEX NUMBERS
- EXAMPLE 1.16 Evaluate (-v/3 - i)^30.
S olution (- v /;; .> - i)^30 = ( 2e' •(-•"))30 --. = 230e-•.^2 5" = -230.
An interesting application of the laws of exponents comes from putt ing theequation (ei^8 )n = ein8 in its polar form. Doing so gives
(cos 8 + i sin O}" = cosn8 + i sin nll, (1-40)
which is known as De Moivre's formula, in honor of the French mathematician
Abraham De Moivre (1667- 1754).
- EXAMPLE 1 .17 Use De Moivre's formula (E.quation (1-40)) to show that
cos 50 = cos^5 0 - 10 cos^3 Osin^2 0 + 5 cos Osin^4 0.
S olut ion If we let n = 5 and use the binomial formula to expand the left side
of Equation (1-40), we obtain
cos^5 0 + i5cos^4 8sin0 - 10cos^3 Osin^2 8 - 10icos^2 Osin^3 0 + 5cos8sin^4 8 + isin^5 8.
The real part of this expression is cos^5 8 - 10cos^3 8sin^2 0+5cos8 sin^4 8. Equating
this to the real part of cos 58 + isin 58 on the right side of Equation (1-40)
establishes the desired result.
A key aid in determining roots of complex numbers is a corollary to the
fundamental theorem of algebra. We prove this theorem in Chapter 6. Our
proofs must be independent of the conclusions we derive here because we are
going to make use of the corollary now.
• EXAMPLE 1.18 Let P (z) = z^3 + (2 - 2i) z^2 + (- 1 -4i) z -2. This polyno-
mial of degree 3 can be written as P (z) = (z - i)^2 (z + 2). Hence the equation
P (z) = 0 has solutions z 1 = i, z2 = i, and z3 = -2. Thus, in accordance with
Theorem 1.4, we have three solutions, with z 1 and z2 being repeated roots.